Segre Embedding

Throughout this article, k is a fixed algebraically closed field. We wish to construct the product in the category of quasi-projective varieties.

For our first example, let be the projective variety defined by the homogeneous equation . We define maps as follows

Note that the maps are well-defined: if then since we have .

Proposition 1.

The triplet is a product in the category of quasi-projective varieties.

Proof

Let be a quasi-projective variety and be morphisms. We will define the corresponding as follows. For each , there is an open neighbourhood U of w such that and where are homogeneous of the same degree and either or has no zero in U. Same holds for .

Now define by . Clearly the image of f lies in V so we get a morphism . It is easy to see that and . Repeating this construction over an open cover of W, we obtain our desired . ♦

Using similar techniques, we can show the following.

Proposition 2.

For any , the product exists in the category of quasi-projective varieties and is a projective variety.

Specifically, the product is the image of the Segre embedding

,

where the projective coordinates of are indxed by with and .

We denote the image of this map by .

Exercise A

Prove that is the closed subspace of defined by

over all .

Products of Quasi-Projective Varieties

Proposition 3.

If and are open (resp. closed), so is the image of in .

In particular, the topology on is at least as fine as the product topology.

Proof

It suffices to prove the case where and are open. Pick any and ; without loss of generality say .

Since is open in there exists a homogeneous such that . Similarly, there exists a homogeneous such that . Then

so the image of in V is open. ♦

As in the product of affine varieties, the topology on is in general strictly finer than the product topology. This is already clear in the case m = n = 1, since has the cofinite topology.

Corollary 1.

The product of two projective (resp. quasi-projective) varieties exists and is projective (resp. quasi-projective).

Note

In the following proof, we say a subset of a topological space is locally closed if it is an intersection of an open subset and a closed subset. Thus every quasi-projective variety (resp. quasi-affine variety) is a locally closed subspace of some (resp. ).

Prove the following properties as a simple exercise:

an intersection of two locally closed subsets is locally closed;

if Y is a locally closed subset of X and Z is a locally closed subset of Y then Z is a locally closed subset of X;

a subset Y of X is locally closed if and only if Y is open in its closure in X.

Proof

If and are closed (resp. locally closed), so is the image W of in by proposition 3. The projections and then restrict to and .

Let us show that is the product of and in the category of quasi-projective varieties.

If X is any quasi-projective variety and , are any morphisms then and induce ; the image of f lies in W so we obtain an induced . ♦

Exercise B

1. Let be the set of points satisfying . Find a set of homogeneous polynomials in which define the image of W.

2. More generally prove that a subset is closed if and only if its corresponding subset is the set of solutions of some bihomogeneous polynomials

,

i.e. F is homogeneous as a polynomial in as well as .

Dimensions

Lemma 1.

For any point in a quasi-projective variety V, there is an open neighbourhood U, , which is affine.

Proof

Suppose is a locally closed subset. Without loss of generality, so is contained in , a locally closed subset of . Now W is open in , its closure in . By an analogue of proposition 1 here, we can pick a basis of the topological space in the form of , where

Thus for some we have . Now we are done since is isomorphic to the affine variety with coordinate ring . ♦

Exercise C

Prove that if V and W are irreducible quasi-projective varieties, then is also irreducible. Again, please be reminded that is not the product topology.

Proposition 4.

if V and W are quasi-projective varieties, then

.

Proof

Suppose V and W are irreducible; by lemma 1 we can pick open affine subsets and . Then is an open affine subset of the quasi-projective variety , which is irreducible by exercise C. Now

Serre’s Criterion for Normality

Throughout this article, fix an algebraically closed field k.

In this section, A denotes a noetherian domain. We will describe Serre’s criterion, which is a necessary and sufficient condtion for A to be normal. In the following section, we will relate the results here to an interesting example from the last article.

Lemma 1.

If A is normal, then for all , we have

Note

Since A is a domain, it has only one minimal prime: 0. Hence all associated primes of A/aA have height at least 1. Lemma 1 thus says principal ideals of a normal domain have no embedded primes.

Proof

We need to show has height 1. Pick such that has annihilator , i.e. . Since bA is finitely generated, we localize both sides at to obtain , the unique maximal ideal of . Thus

and . We claim that ; if not, and by the adjugate matrix trick (see proof of proposition 6 here), is integral over B. This contradicts the fact that B is normal. Hence is an invertible ideal so B is a dvr, and . ♦

Lemma 2.

Suppose for all , , we have . Then

,

where intersection occurs in .

Proof

Let , where , ; we need to show . Write for its primary decomposition with associated primes all of height 1. For each i we have . But since all are minimal in . Thus

Let for an irreducible affine variety V, and be a prime ideal with corresponding subvariety . Prove that is the set of all rational functions on V which are regular at some point of W.

Note

We already know holds for all domains; geometrically, this means if (for irreducible V) is regular at each point, then f can be represented by the same polynomial globally. The condition in lemma 2 is notably stronger; geometrically, it says if f is regular on an open dense subset of every codimension 1 subvariety, then it is regular everywhere.

Theorem (Serre’s Criterion).

A noetherian domain is normal if and only if the following conditions both hold.

All , for , have height 1.

For each of height 1, is a dvr.

When that happens, .

Note

In the context of algebraic geometry, the first condition says “subvarieties cut out by a single equation have no embedded components” while the second says “the set of singular points has codimension at least 2” (this will be elaborated in later articles). Thus normality can fail in two different ways: hidden (embedded) components or too many singular points.

Let be a chain of irreducible closed subsets of Y. Taking their closures in X we have

.

By proposition 2 here, each is a closed irreducible subset of X. Furthermore by a general result in point-set topology (see proposition 3 here), the closure of any subset in Y is . Hence so for any i. Thus we get a chain of irreducible closed subsets of X of length k. ♦

To proceed, we need the following correspondence.

Proposition 2.

Let U be a non-empty open subset of a topological space X. Then there is a bijective correspondence:

(⇒) Suppose V is irreducible; let . If are homogeneous, then and are closed subsets of properly contained in V. Since V is irreducible the following shows :

.

(⇐) Let where is prime. Let be closed subsets with union V. Now write and for homogeneous radical ideals and . Then . Since is a homogeneous radical ideal, . By exercise B here, or . ♦

Corollary 1.

Let be a non-empty closed subset. Then is irreducible if and only if is irreducible.

Proof

By proposition 1, V is irreducible if and only if is prime. But (from an exercise here) so the result follows. ♦

Quasi-projective Varieties

Recall that a projective variety is a closed subset of some .

Definition.

A quasi-projective variety is an open subset of a projective variety.

This merely defines it as a set; we need a geometric structure on it.

First, let be homogeneous polynomials of the same degree. If is such that not all , then we can define a function on an open subset U of containing as follows:

.

The map is well-defined: indeed if we can find an open neighbourhood U of such that . Also, if we replace projective coordinates with , then each where so

We write for the resulting function.

Definition.

Let and be quasi-projective varieties and be a function.

We say is regular at if there is an open neighbourhood U of in V such that

for some homogeneous of the same degree.

We say is regular if it is regular at every , in which case we also say is a morphism of quasi-projective varieties.

From the above definitions, we obtain the category of all quasi-projective varieties and morphisms between them.

Example 1

First consider the case where and are closed subsets.

E.g., let . A regular map in the earlier sense can be expressed as a polynomial , e.g. take . Via embeddings and taking and respectively, f can be written in terms of homogeneous coordinates as

since it is the homogenization of the map . This generalizes to an arbitrary regular map of closed subsets .

Conversely we have:

Lemma 2.

Let be regular under the new definition. Then there exist polynomials which represent .

Proof

We will prove this for the case where V is irreducible.

For each of , let be projection onto the i-th coordinate. Then is regular under the new definition, and by proposition 2 here (and its preceding discussion) can be represented as a polynomial . Hence we see that

for polynomials . ♦

Example 2

Take the map given by

Note that the same set of polynomials works globally over the whole of .

Example 3

Suppose . Let be the closed subset defined by . We define a map as follows

Outside the point (1 : 1 : 0), take .

Outside the point (1 : -1 : 0), take .

The map agrees outside those two points since due to the equality .

Isomorphisms

Definition.

Consider the category of all quasi-projective k-varieties, with morphisms defined as above. Two such varieties are said to be isomorphic if they are isomorphic in the category.

A quasi-projective variety is said to be

projective if it is isomorphic to a closed subset of some (this generalizes the existing definition of projective varieties);

affine if it is isomorphic to an affine k-variety (closed subspace of some );

quasi-affine if it is isomorphic to an open subset of an affine variety.

Example 4

In example 3 above, we get an isomorphism since we have the reverse map

.

As an exercise, prove that and .

Definition.

The coordinate ring of a quasi-projective variety V is the set

,

taken to be a k-algebra via point-wise addition and multiplication:

Note

As before, a regular map of quasi-projective varieties induces a ring homomorphism . By lemma 2, when V is affine agrees with our earlier version (we proved this in the case where V is irreducible).

Example 5

For each , we have an automorphism

Note that for . Also if and only if g is a scalar multiple of the identity matrix, so we get an injective homomorphism . In fact this is an isomorphism of groups.

E.g. when n = 1, we get the Möbius transformations:

Example 6

We have an isomorphism between the quasi-affine variety and via the maps

Hence is an affine variety even though it is not closed in . From the isomorphism we also have:

Example 7

Let . We will show that V is not affine. Indeed consider the injection which induces

.

The map is injective since V is dense in . Let us show that it is surjective. Suppose so that is regular. Write

where .

By example 6, we have and . Since are all dense in V we have injections and so that . It is easy to show that this means .

Hence induces an isomorphism of the coordinate rings . If V is affine, by proposition 1 here would be an isomorphism of varieties, which is a contradiction since is not surjective.

For a finite sequence of homogeneous polynomials we also write for where .

Also let .

Exercise A

Prove the following, for any graded ideals and collection of graded ideals of B.

.

If is a set of homogeneous generators of , then

.

.

.

In the other direction, we define:

Definition.

Let be any subset. Then denotes the (graded) ideal of B generated by:

.

In summary, we defined the following maps.

Zariski Topology of Projective Space

We wish to define the Zariski topology on ; for that let us take subsets of which can be identified with the affine space . Fix ; let

Note that for the same point can be represented by where the i-th coordinate is 1. This gives a bijection . E.g. for n = 2, we have:

Note that for any , the intersection maps to an open subset of via both and . Indeed if i < j then is the set of all satisfying while is the set of all satisfying . Hence, the following is well-defined.

Definition.

The Zariski topology on is defined by specifying every as an open subset, where obtains the Zariski topology of from .

A projective variety is a closed subspace .

First, we have the following preliminary results.

Lemma 1.

For any homogeneous , the set is (Zariski) closed in .

Proof

It suffices to show that is closed in for each . But , where . The same holds for . ♦

Definition.

For any , let

be its homogenization.

Exercise B

1. Prove that if , the homogenization of fg is the product of the homogenizations of f and g.

2. Let be the homogenization of a non-constant . Then f is irreducible if and only if F is irreducible. [ Hint: you may find lemma 2 here helpful. ]

The Zariski topology on is consistent with our earlier notions of closed subsets:

Proposition 1.

A subset is closed under the Zariski topology if and only if for some graded ideal .

Proof

(⇐) Suppose for some homogeneous . Since , by lemma 1 this is Zariski closed.

(⇒) Let ; it suffices to show that any is contained in for some homogeneous . Now is contained in some , say without loss of generality. Hence for some . If , then

, where F = homogenization of f. ♦

Example

Let be the projective variety defined by the homogeneous equation . Then

is cut out from by ;

is cut out from by ;

is cut out from by .

Cone of Projective Variety

We wish to prove the bijective correspondence between graded radical ideals of B and closed subsets . For that, we can piggyback on existing results for the affine case.

Definition.

Let be any subset. The cone of C is

.

Note

For any non-empty collection of subsets we have

Also, we have:

Lemma 2.

If is a proper graded ideal then .

In particular, by proposition 1, the cone of a closed is closed in .

Proof

Note: if is non-constant homogeneous, then . Now pick a set of homogeneous generators for ; each is non-constant so

.

This completes the proof. ♦

Furthermore we have:

Lemma 3.

A non-empty closed subset is of the form for some closed if and only if

.

When that happens, we call W a closed cone in .

Proof

(⇒) is obvious; for (⇐) clearly W = cone(V) for some subset . Let so that . It remains to show is graded, for we would get by lemma 2.

Indeed if , write as a sum of homogeneous components. Then for any and we have which gives

Thus vanish for any , i.e. . ♦

Projective Nullstellensatz

Thus we have the following correspondences:

The top-left column is a bijection by lemma 3; the bottom row is a bijection by Nullstellensatz. In the proof of lemma 3, we also showed that for a closed cone , the ideal is graded. Conversely, if is graded, is the (non-empty) solution set of a collection of graded polynomials; hence it is a closed cone too.

Hence we have a bijection between

closed subsets , and

proper homogeneous radical ideals .

The correspondence takes and so

The last piece of the puzzle is the map which takes closed subsets of to homogeneous ideals of B. As an easy exercise, show that

.

However so we modify our bijection to:

Theorem (Projective Nullstellensatz).

There is a bijection between:

closed subsets ;

homogeneous radical ideals such that where is the irrelevant ideal of B.

Exercise C

Prove that if is a homogeneous ideal then is empty if and only if contains a power of .

(⇒) Suppose . By proposition 3 here, the set of zero-divisors (in A) of is . Now if and , then is a zero-divisor of so .

To prove that , suppose ; then is a zero-divisor for so and we have . Conversely, if , then since every minimal prime in is an associated prime of (by proposition 2 here), x is contained in every minimal prime of . By proposition 5 here, .

(⇐) Suppose satisfies the given condition; we first show that is prime. Suppose and . For some n > 0, . But so by the given condition and .

Let ; it remains to show . First, is the unique minimal element of so .

Conversely, it remains to show any zero-divisor of as an A-module lies in . But if and are such that , then by the given condition . ♦

We thus say a proper ideal is primary if:

.

Note that prime ideals are primary, and by proposition 1, the radical of a primary ideal is prime.

Exercise A

1. Prove that if is a ring homomorphism and is a primary ideal, then is a primary ideal of A. Also .

2. Prove that for an ideal , there is a bijection between primary ideals of A containing and primary ideals of .

A power of a prime ideal is not primary in general. E.g. let k be a field and with , which is prime since

.

Then is not primary because but and .

Worked Examples

Throughout this section, k denotes a field.

Example 1

In an earlier example, we saw that for and , the A-module has two associated primes: (X) and (X, Y). The following are primary decompositions, both of which are clearly minimal:

.

To check that these ideals are primary:

is prime and hence primary;

is a power of the maximal ideal ; by exercise B.2 it is primary;

so is primary by exercise B.2.

Note that . Geometrically, the k-scheme with coordinate ring looks like the following.

Example 2

Let with . Then

.

We have , an intersection of primary ideals with and . This translates to

with and is already prime.

Example 3

Let with . Then each of , and is primary (the first ideal is prime; the remaining two ideals are powers of a maximal ideal). Hence this gives a primary decomposition of so its associated primes are , and , with the latter two embedded.

Geometrically, the k-scheme with coordinate ring looks like:

Computing an explicit set of generators for is not trivial, but it can be done with Buchberger’s algorithm.

Prime Composition Series

Throughout this article, A is a noetherian ring and all A-modules are finitely generated.

Recall (proposition 1 here) that if M is a noetherian and artinian module, we can find a sequence of submodules whose consecutive factors are simple modules. Correspondingly we have:

Proposition 1.

For finitely generated M, there exists a sequence of submodules

such that each as A-modules for prime ideals .

Proof

Assume . Since by proposition 3 here, there is an embedding of A-modules . If equality holds, we are done. Otherwise, let be the image of the map and repeat with to obtain a submodule . Repeating this process, this must eventually terminate since we cannot have an infinite ascending chain of submodules . ♦

Let k be a field, and , considered as an A-module. Note that with radical so we get (by proposition 1 here)

.

On the other hand, we claim that . Indeed we have:

so and is an embedded prime of M.

Conversely, we pick the chain of submodules with generated by . Then ; as shown above, . Also so by corollary 1

.

Exercise A

Find an A-module M and prime ideals with but .

Existence of Primary Decomposition

In this section we fix an ambient non-zeroA-module M (finitely generated of course) and consider its submodules. For each prime , take the set of all submodules such that . Note that

.

Now for each prime , fix a maximal element .

Lemma 2.

We have .

Note

If then so the above intersection only needs to be taken over , i.e. a finite number of terms.

Proof

Let . If it has an associated prime . Since we have , a contradiction. ♦

Lemma 3.

Let . Then .

Proof

By proposition 4 here, we have . Since we have . On the other hand if , then we have an injection whose image is of the form . But now

which does not contain , contradicting maximality of . Hence no such exists. ♦

Definition.

For an associated prime of M, a –primary submodule of M is an such that .

A primary decomposition of M is an expression

where each is a -primary submodule of M.

The decomposition is irredundant if for any , . It is minimal if it is irredundant and all are distinct.

By lemmas 2 and 3, a minimal primary decomposition exists for every non-zero module.

It is not true that in every primary decomposition , must be a maximal element of . We will see an example in the next article.

Exercise B

Prove that if are -primary submodules, so is .

Hence given any primary decomposition, we can get a minimal one by first removing the redundant terms then taking the intersection of all with the same corresponding .

Properties of Primary Decomposition

Here, we will discuss properties of a general primary decomposition. Throughout this section we fix:

where is -primary in M.

Proposition 3.

Every must occur among the .

Proof

Since the canonical map is injective. Hence

♦

Proposition 4.

If the primary decomposition is irredundant, then every is an associated prime of M.

Proof

If then among the injections and , we have since and are disjoint. Thus

is injective

and , contradicting the condition of irredundancy. ♦

Note

Thus proposition 4 gives us a way to compute : find a primary decomposition and remove terms until it becomes irredundant. However, note that proposition 4 does not require to be distinct.

Corollary 2.

If the primary decomposition is minimal, then

has exactly n elements.

Proof

Apply propositions 3 and 4. ♦

Finally, we define the primary decomposition of submodules.

Defintion.

Let be a submodule. A primary decomposition of N in M is an expression

such that is a -primary submodule of M.

Note

A primary decomposition of N in M corresponds bijectively to a primary decomposition of M/N. We say the primary decomposition of N in M is irredundant or minimal if the corresponding primary decomposition of M/N is so.

In a UFD, every non-zero element is a unique product of irreducible (also prime) elements.

In a Dedekind domain, every non-zero ideal is a unique product of maximal ideals.

Here, we will introduce yet another type of factorization, called primary decomposition. The main idea is that in a noetherian ring, every ideal (even the zero ideal) is an intersection of primary ideals. E.g. in , every non-zero ideal is an intersection of , ideals generated by prime powers.

Annihilators

Let A be a fixed ring and M be an A-module.

Definition.

The annihilator of in A is

,

an ideal of A. Similarly, the annihilator of M in A is

,

also an ideal of A. If , we say M is a faithful A-module.

Note that if , then M can be regarded as a faithful -module.

Exercise A

1. Given A-submodules , let

.

State and prove the analogue of proposition 2 here for submodules of M. Observe that we can write the annihilators as

Conversely, express in terms of annihilators.

2. Prove that if P is a finitely generated submodule of M.

In particular if M is finitely generated then

.

Support of a Module

Definition.

The support of an A-module M is:

Geometrically, these are points in Spec A at which the module does not vanish.

Proposition 1.

If M is a finitely generated A-module, then

In particular, the support of M is a closed subspace of .

Proof

Let . For we have is zero if and only if there exists such that , equivalently if there exists . Hence if and only if .

In the general case, let generate M as an A-module. Then

For the last equivalence, recall that contains if and only if it contains either or . ♦

Proposition 2.

If is a short exact sequence of A-modules, then .

If N, P are finitely generated A-modules, then

.

Note

Philosophically, if we imagine the first case as N + P and the second case as N × P, then the result says if and only if both terms are zero whereas if and only if at least one term is zero.

Proof

Let be prime.

For the first claim, we have a short exact sequence and it follows that if and only if .

For the second claim, we have

If or , clearly the RHS is zero. Conversely if both are non-zero, since they are finitely generated -modules, Nakayama’s lemma gives and . But these are vector spaces over so we have

♦

Exercise B

1. Let be a homomorphism of finitely generated A-modules; prove that is a closed subset of Spec A.

2. Prove: if is multiplicative then

Associated Primes

Definition.

Let be a prime ideal. We say is associated to the A-module M if for some .

Let be the set of prime ideals of A associated to M.

Note

We have: if and only if there is an injective A-linear map . Observe that if the annihilator of is , then so is that of any non-zero multiple of m; this follows immediately from the definition of prime ideals.

Next suppose M is an -module; we can compute both and . There is a bijection

.

In particular, we can compute by considering as an A-module or a module over itself. The above shows that there is effectively no difference.

Proposition 3.

Suppose A is noetherian. The union of all associated primes of M is its set of zero-divisors

.

In particular, any non-zero A-module M has an associated prime.

Proof

Fix an and such that ; we need to show a is contained in an associated prime of M.

So let be the set of ideals of A containing a of the form for . Since A is noetherian and , there is a maximal . We will show is prime; first write for some .

Pick such that and . Since , we have . And since , by maximality of we have so since we have . ♦

Henceforth, A denotes a noetherian ring. However, we do not assume all modules are finitely generated at first.

Properties of Associated Primes

Proposition 4.

Let be a short exact sequence of A-modules. Then

Proof

We may assume is a submodule and . The first containment in the claim is obvious. For the second, suppose we have an A-linear map .

If , then any non-zero has annihilator so . If , then composing is still injective, so . ♦

Corollary 1.

We have .

Proof

Follows immediately from proposition 4. ♦

Next, we show that taking the set of associated primes commutes with localization.

Proposition 5.

Let be a multiplicative subset. Then

Proof

(⊇) If and , then localization at S gives an A-linear map .

(⊆) Suppose we have an -linear map where is prime. By theorem 1 here, we can write for some prime such that . Now in the injection we let be the image of 1. It remains to show: there exists such that .

For each , we have in so that in M for some , i.e. . Pick a generating set of ; then for each i there exists with . Then satisfies .

Conversely if then so lies in the annihilator of , i.e. in . Then ; since we have . ♦